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I heard a talk recently on number theoretic representation theory, in which the speaker suggested that we focus on the case $G=$GL$_2$, with $\rho$, the representations, being thought of as sym$^n$ for simplicity.

This is all kind of new to me, so I'm rather confused. In this kind of generality and without an ability to furnish more context than I've already given, is it clear what GL$_2$ is? I guess one fixes some kind of a field of coefficients, but I'm not sure.

And more mysterious to me is what exactly the above representation is. Is this standard notation? From what I gather it has something to do with the symmetric power of a representation, but then it seems to me one would have to have an underlying representation first before one can take symmetric powers of it. Or am I on the wrong path completely and is sym$^n$ some kind of a specific representation of GL$_2$?

I hope this is an appropriate question.

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If it helps the talk had something to do with Langlands program. It was mostly over my head, but I would like to take a few things away from the talk if possible, however small. – Thelonius Sep 15 '12 at 19:35
Well, $\mathrm{GL}_2 (k)$ is a matrix group, so it has a tautological $2$-dimensional representation over $k$... – Zhen Lin Sep 16 '12 at 4:20
usually $\mathrm{GL}_2$ refers to the algebraic group $GL_2(k)$ with $k$ finite. Sometimes $k$ may be infinite and even characteristic zero, depends on the context of the talk this is probably clear to audience who study these kinds of thing. $\mathrm{sym}^n$ is usually $\mathrm{sym}^n(V)$ where $V$ is the standard (2-dimensional) representation for which $\mathrm{GL}_2$ acts as the usual matrix multiplying on vectorspace. It is also possible that they are talking about $\mathrm{sym}^n(V)$ for others representation $V$ but I guess this is unlikely. – Aaron Sep 20 '12 at 12:06
@Aaron Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer May 27 '14 at 9:32

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